Unit Revision

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HSN23510

Unit 3 Level C AssessmentThese notes were created specially for the Higher Still Notes website, and we require that any copies or derivativeworks attribute the work to us.

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Higher

Mathematics


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Notes

Higher Mathematics Unit 3 Level C Assessment

Page 1 www.hsn.uk.net

HSN23510

Outcome 1

1. a) A, B and C have co-ordinates ( )4, 3,1 , ( )0, 1,0 and ( )4,1, 1 respectively.(i) Write down the components ofAC

.

(ii) Hence show that the points A, B and C are collinear. 4

b) The point R divides the line ST in the ratio 3 : 2 , as shown in the diagram.

Find the co-ordinates of R. 3

2. The diagram shows the triangle LMN where3

LM 4

2

=

and

2

LN 4

5

=

a) Find the value of LM.LN

. 1

b) Use the result of (a) to find the size of angle MLN. 4

Outcome 2

3. a) Differentiate 2sinx with respect tox. 1b) Given 5cosy x= , find

dy

dx

. 1

4. Find ( )f x when ( )13( ) 2 7f x x= + . 25. a) Find 3 cos

2x dx . 2

b) Integrate 3sinx with respect tox. 1

c) Evaluate ( )36

43x dx . 4

L

N

M

R

( )T 7, 5,1

..

.

( )S 3,10,6


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Outcome 3

6. a) Simplify log 7 log 3a a+ . 1b) Simplify 3 3log 5 3log 2 . 3

c) Simplify 2log 2 . 1

7. a) If log 7log 4

e

e

x= , find an approximation forx. 1

b) Given that10

log 3.1y= , write an expression for the exact value ofy. 1

c) If 2.910y= , find an approximation fory. 1

Outcome 4

8. Express 12 cos 5sinx x + in the form ( )cosk x a where 0k> and 0 360a . 5

Total Marks: 25

[END OF QUESTIONS]


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HSN23510

Pass Marks

Outcome 1 Outcome 2 Outcome 3 Outcome 4

9

12

8

11

5

8

3

5

Marking Scheme

Outcome 1 Vectors

1 a i AC

4 4

1 3

1 18

4

2

4

2 2

1

=

=

=

=

c a

Calculate AC

1

ii AB

0 4

1 3

0 1

4

2

1

=

=

=

b a

Since 2AB AC=

and A is a common point,A, B and C are collinear (or any other appropriate statement)

Know to calculate AB Calculate directed line

segment

Statement

3


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HSN23510

b

( ) ( )

( )

SR 3

2RT

2SR 3RT

2 3

2 2 3 35 3 2

13 2

5

7 31

3 5 2 105

1 6

21 6115 20

53 12

151

55

15

3

13

=

=

=

= = +

= +

= +

= +

+

=

=

r s t r

r s t r r t s

r t s

( )R 3,1,3 (or Section Formula)

State ratioRearrange to give rState co-ordinate of R

3

2 a ( ) ( ) ( )LM.LN 3 2 4 4 2 5

6 16 10

20

= + +

= + +

=

Calculate scalarproduct

1

b

( )22 2 2 2 2

LM.LN

cos LM LN

20

3 4 2 2 4 5

0.554

56.4

=

=

+ + + +

=

=

StrategySubstitute valuesState cosProcess angle

4

Outcome 2 Further Calculus

3 a

( )2sin 2cos

d

x xdx =

Differentiate correctly1

b5sin

dyx

dx=

Differentiate correctly1


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HSN23510

4 ( ) ( )132 7f x x= +

( ) ( )

( )

( )2

3

23

23

12 7 2

3

22 7

3

2

3 2 7

f x x

x

x

= +

= +

=

+

Correct multiplyingfactor (ie 23 )

Bracket correct andraised to appropriatepower

2

5 a 3 3cos sin

2 2x dx x C= +

IntegrateAdd constant of

integration 2

b 3sin 3cosx dx x C= + Integrate 1c ( )

( )

( ) ( )

6 3

4

4

4 4

4 4

6

4

3

3

4

6 3 4 3

4 4

3 1

4 4

80 4

=20

x dx

x

=

=

=

=

Bracket correct andraised to appropriatepower

Correct multiplyingfactor (ie 14 )

Substitute limitsCalculate integral

4

Outcome 3 Exponentials and Logarithms

6 a log 7 log 3 log 21a a a+ = Apply add rule 1b 3

3

3 3 3 3

3

3

log 5 3log 2 log 5 log 2

5log

2

5log

8

=

=

=

Apply power ruleApply subtract ruleState solution

3

c 2log 2 1= Interpret loga a 17 a log 7

log 4

1.404

e

e

x=

=

Process solution

1

b3.1

10log 3.1

10

y

y

=

=

Interpret loga b c= 1

c 2.910

794.3

y=

=

Process solution1


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Outcome 4 The Wave Function

8 ( )cos cos cos sin sink x a k a x k a x = +

cos 12

sin 5

k a

k a

=

=

5tan

12

22.6

a

a

=

=

2 212 5

13

k= +

=

( )12cos 5sin 13cos 22.6x x x + =

Know to expandState cosk a and

sink a Calculate kState tana Calculate a

5

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